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3 years ago

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You place a balloon in a closed chamber at standard temperature and pressure (STP). You cool the chamber to half of its original

temperature while leaving the pressure constant. What happens to the balloon?
It stays the same volume.

It decreases to one-half of its original volume.

It doubles in volume.

It expands to four times its original volume.

2 answers:

Goryan [66]3 years ago

7 0

It decreases to one half of its original volume.

AnnyKZ [126]3 years ago

6 0

I think the correct answer from the choices listed above is the second option. At constant temperature, the volume of the system at the new conditions will decrease to <span>one-half of its original volume.

V1 = nRT1 / P

V2 = nRT1 / 2P = 2V1</span>

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Annette [7]

The phase diagram of CO2 has a melting curve that slopes up and to the right, in contrast to the phase diagram of water, which has a more conventional shape. It is impossible for liquid CO2 to exist at pressures lower than 5.11 atm because the triple point is 5.11 atm and 56.6 °C.

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1 year ago

How many moles are in 8 grams of NaCl?

anygoal [31]

There are 2 moles in 8 grams

5 0

3 years ago

Do this Q7 if someone do I will her or him brainliest

Simora [160]

The average speed :

1. 10.44 m/s

2. 10.42 m/s

3. 9.26 m/s

The distance 100 m have the greatest average speed

<h3>Further explanation </h3>

Given

Distance and time of runner

Required

Average speed

Solution

<em> Average speed : total distance : total time </em>

1. d = 100 m, t = 9.58 s

Average speed : 100 : 9.58 = 10.44 m/s

2. d=200 m, t=19.19 s

Average speed : 200 : 19.19 = 10.42 m/s

3. d=400 m, t = 43.18 s

Average speed : 400 : 43.18 = 9.26 m/s

The distance 100 m have the greatest average speed

7 0

3 years ago

I need someone to answer these.

SCORPION-xisa [38]

Answer:

The identity of an atom is determined my the number of <u>protons</u>. This is the <u>atomic number</u>.

The particle(s) found inside the nucleus are called <u>protons and neutrons</u>. Their combined mass is referred to as <u>the mass number</u>.

Isotopes have the same number of <u>protons</u>, but different number of <u>neutrons</u>.

7 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

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